1 String Matching Algorithms Topics  Basics of Strings  Brute-force String Matcher  Rabin-Karp String Matching Algorithm  KMP Algorithm

CSE5311 Kumar 2 In string matching problems, it is required to find the occurrences of a pattern in a text. These problems find applications in text processing, text-editing, computer security, and DNA sequence analysis. Find and Change in word processing Sequence of the human cyclophilin 40 gene CCCAGTCTGG AATACAGTGG CGCGATCTCG GTTCACTGCA ACCGCCGCCT CCCGGGTTCA AACGATTCTC CTGCCTCAGC CGCGATCTCG : DNA binding protein GATA-1 CCCGGG : DNA binding protein Sma 1 C: Cytosine, G : Guanine, A : Adenosine, T : Thymine

CSE5311 Kumar 3 Text : T[1..n] of length n and Pattern P[1..m] of length m. The elements of P and T are characters drawn from a finite alphabet set . For example  = {0,1} or  = {a,b,..., z}, or  = {c, g, a, t}. The character arrays of P and T are also referred to as strings of characters. Pattern P is said to occur with shift s in text T if 0  s  n-m and T[s+1..s+m] = P[1..m] or T[s+j] = P[j] for 1  j  m, such a shift is called a valid shift. The string-matching problem is the problem of finding all valid shifts with which a given pattern P occurs in a given text T.

CSE5311 Kumar 4 Brute force string-matching algorithm To find all valid shifts or possible values of s so that P[1..m] = T[s+1..s+m] ; There are n-m+1 possible values of s. Procedure BF_String_Matcher(T,P) 1. n  length [T]; 2. m  length[P]; 3. for s  0 to n-m 4.do if P[1..m] = T[s+1..s+m] 5. then shift s is valid This algorithm takes  ((n-m+1)m) in the worst case.

CSE5311 Kumar 5 a ca ab c a ab aa b acaabc aab acaabcmatches aab

CSE5311 Kumar 6 Rabin-Karp Algorithm Let  = {0,1,2,...,9}. We can view a string of k consecutive characters as representing a length-k decimal number. Let p denote the decimal number for P[1..m] Let t s denote the decimal value of the length-m substring T[s+1..s+m] of T[1..n] for s = 0, 1,..., n-m. t s = p if and only if T[s+1..s+m] = P[1..m], and s is a valid shift. p = P[m] + 10(P[m-1] +10(P[m-2] (P[2]+10(P[1])) We can compute p in O(m) time. Similarly we can compute t 0 from T[1..m] in O(m) time.

CSE5311 Kumar 7 p = P[m] + 10(P[m-1] +10(P[m-2] (P[2]+10(P[1])) 6378 =    10 3 = ( (3 + 10(6))) = m =4

CSE5311 Kumar 8 t s+1 can be computed from t s in constant time. t s+1 = 10(t s –10 m-1 T[s+1])+ T[s+m+1] Example : T = t s = 31415, s = 0, m= 5 and T[s+m+1] = 2 t s+1 = 10(31415 –10000*3) +2 = Thus p and t 0, t 1,..., t n-m can all be computed in O(n+m) time. And all occurences of the pattern P[1..m] in the text T[1..n] can be found in time O(n+m). However, p and t s may be too large to work with conveniently. Do we have a simple solution!!

CSE5311 Kumar 9 Computation of p and t 0 and the recurrence is done using modulus q. In general, with a d-ary alphabet {0,1,…,d-1}, q is chosen such that d  q fits within a computer word. The recurrence equation can be rewritten as t s+1 = (d(t s –T[s+1]h)+ T[s+m+1]) mod q, where h = d m-1 (mod q) is the value of the digit “1” in the high order position of an m-digit text window. Note that t s  p mod q does not imply that t s = p. However, if t s is not equivalent to p mod q, then t s  p, and the shift s is invalid. We use t s  p mod q as a fast heuristic test to rule out the invalid shifts. Further testing is done to eliminate spurious hits. - an explicit test to check whether P[1..m] = T[s+1..s+m]

CSE5311 Kumar 10 t s+1 = (d(t s –T[s+1]h)+ T[s+m+1]) mod q h = d m-1 (mod q) Example : T = 31415; P = 26, n = 5, m = 2, q = 11 p = 26 mod 11 = 4 t0 = 31 mod 11 = 9 t1 = (10(9 - 3(10) mod 11 ) + 4) mod 11 = (10 (9- 8) + 4) mod 11 = 14 mod 11 = 3

CSE5311 Kumar 11 Procedure RABIN-KARP-MATCHER(T,P,d,q) Input : Text T, pattern P, radix d ( which is typically =  ), and the prime q. Output : valid shifts s where P matches 1. n  length[T]; 2. m  length[P]; 3. h  d m-1 mod q; 4. p  0; 5. t 0  0; 6. for i  1 to m 7. do p  (d  p + P[i] mod q; 8. t 0  (d  t 0 +T[i] mod q; 9. for s  0 to n-m 10. do if p = t s 11. then if P[1..m] = T[s+1..s+m] 12. then “pattern occurs with shift ‘s’ 13. if s < n-m 14. then t s+1  (d(t s –T[s+1]h)+ T[s+m+1]) mod q;

CSE5311 Kumar 12 Comments on Rabin-Karp Algorithm qAll characters are interpreted as radix-d digits qh is initiated to the value of high order digit position of an m-digit window qp and t 0 are computed in O(m+m) time qThe loop of line 9 takes  ((n-m+1)m) time The loop 6-8 takes O(m) time The overall running time is O((n-m)m)

CSE5311 Kumar 13 Exercises -- Home work Study KMP Algorithm for String Matching -- Knuth Morris Pratt (KMP) Study Boyer-Moore Algorithm for String matching Extend Rabin-Karp method to the problem of searching a text string for an occurrence of any one of a given set of k patterns? Start by assuming that all k patterns have the same length. Then generalize your solution to allow the patterns to have different lengths. Let P be a set of n points in the plane. We define the depth of a point in P as the number of convex hulls that need to be peeled (removed) for p to become a vertex of the convex hull. Design an O(n 2 ) algorithm to find the depths of all points in P. The input is two strings of characters A = a1, a2, …, an and B = b1, b2, …, bn. Design an O(n) time algorithm to determine whether B is a cyclic shift of A. In other words, the algorithm should determine whether there exists an index k, 1  k  n such that ai = b(k+i) mod n, for all i, 1  i  n.